Physics

Electromagnetic Waves and Spectrum

659 Questions

Electromagnetic waves and spectrum questions test a candidate's understanding of radiation frequencies, wavelengths, and the properties of different rays like infrared, ultraviolet, and visible light. Concepts also cover practical applications in astronomy and the fundamental speed of light calculations. This physics topic appears regularly in general science sections of major competitive examinations.

UV rays propertiesElectromagnetic radiation frequencyWavelength identificationSpeed of light calculationsBlack body radiation

Electromagnetic Waves and Spectrum Questions

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength $5200\overset {\circ}{A}$?

  1. $700\ m/s$
  2. $1000\ m/s$
  3. $1400\ m/s$
  4. $2800\ m/s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Momentum, $p = mv = \dfrac {h}{\lambda}$


or $v = \dfrac {h}{m\lambda}$

$\therefore v = \dfrac {6.62\times 10^{-34}}{9.1\times 10^{-31}\times 5.2\times 10^{-7}}$

$\Rightarrow v = \dfrac {6.2\times 10^{4}}{9.1\times 5.1}$

$\Rightarrow v = 1400\ m/s$.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Momentum of a photon having frequency $1.5\times 10^{13}Hz$?

  1. $3.13\times 10^{-29}kg m/s$
  2. $3.3\times 10^{-34}kg m/s$
  3. $6.6\times 10^{-34}kg m/s$
  4. $6.6\times 10^{-30}kg m/s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\quad Momentum\quad of\quad a\quad photon\quad is\quad P=\quad h/\lambda \ \quad And\quad \quad \quad \lambda \nu =c\quad ,\quad where\quad c=\quad speed\quad of\quad light\quad h=\quad Planck's\quad constant\ \qquad \qquad \qquad \qquad \qquad \qquad \lambda =\quad wavelength\quad of\quad photon\ \qquad \qquad \qquad \qquad \qquad \qquad \nu =\quad frequency\quad of\quad photon\ so\quad \lambda =\quad \dfrac { 3\times { 10 }^{ 8 } }{ 1.5\times { 10 }^{ 13 } } =2\times { 10 }^{ -5 }{ m }^{ }\quad P=\dfrac { 6.26\times { 10 }^{ -34 } }{ 2\times { 10 }^{ -5 } } =3.13\times { 10 }^{ -29 }kg m/s\ Therefore\quad option\quad A.\quad$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Calculate the number of photons emitted per seconds from a monochromatic light source of $40\ W$, giving out light of wavelength $5000\ \mathring {A}$.

  1. $1.0\ \times 10^{30}$
  2. $1.0\ \times 10^{20}$
  3. $1.0\ \times 10^{10}$
  4. $1.0\ \times 10^{25}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power P = n * E_photon. E_photon = hc / lambda = 12400 eV-A / 5000 A = 2.48 eV = 2.48 * 1.6e-19 J. n = P / E_photon = 40 / (2.48 * 1.6e-19) = 1.0e20.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Momentum of $\gamma-ray$ photon of energy $3\ keV$ in $kg-m/s$ will be

  1. $2.95 \times 10^{-23}$
  2. $1.6\times 10^{-21}$
  3. $1.6\times 10^{-24}$
  4. $1.6\times 10^{-27}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If the energy of the electron is $E=3KeV=3\times 10^3 eV=3\times 1.6\times 10^{-16}Joule$

 
Then the momentum ,$p$ is given by  $p=\sqrt[2]{2mE}=\sqrt[2]{2\times 9.1\times 10^{-31} \times 4.8\times 10^{-16}}=2.95\times 10^{-23} Kgm/s$
Where $m=9.1\times 10^{-31}Kg$ is mass of electron.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Total energy of $electron$ is more than energy of $photon$ if both are having $equal\ \lambda.$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a photon, E = pc. For an electron, E = p^2 / 2m. Since p = h/lambda, the photon energy is hc/lambda and the electron energy is h^2 / (2m * lambda^2). Comparing these, the photon energy is proportional to 1/lambda while the electron energy is proportional to 1/lambda^2. At large lambda, the photon energy is greater.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Monochromatic light of wavelength 440 nm is produced. The power emitted by light is 18 mW, The number of photons emitted per second by light beam is :

$(h=6.6\times { 10 }^{ -34 })$

  1. $2.09\times { 10 }^{ 16 }$
  2. $4\times { 10 }^{ 16 }$
  3. $3\times { 10 }^{ 18 }$
  4. $4\times { 10 }^{ 18 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,


$\lambda=440nm$


$h=6.6\times 10^{-34}$

$I=18m W$

The energy of monochromatic light,

$E=\dfrac{hc}{\lambda}$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^8}{440\times 10^{-9}}$

$E=0.045\times 10^{-17}J$

The number of photon emitted per second by the light beam,

$n=\dfrac{I}{E}$ 

$n=\dfrac{18\times 10^{-3}}{0.045\times 10^{-17}}$

$n=4\times 10^{16}$

The correct option is B.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The momentum of a photon of energy 1MeV, in kg/m/s, will be :

  1. $\displaystyle 10^{-22}$
  2. $\displaystyle 0.33\times 10^{6}$
  3. $\displaystyle 5\times 10^{-22}$
  4. $\displaystyle 7\times 10^{-24}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The momentum of photon is given by


$p = \dfrac{E}{c}$

where, E is the energy of photon(in eV) and c is the velocity of light , $c = 3 \times 10^8 ms^{-1}$.

$p = \dfrac{1 \times 10^{6} \times 1.6 \times 10^{-19}}{3 \times 10^8}$


$p =  0.533 \times 10^{-21}$

$p = 5 \times 10^{-22} kgsm^{-1}$

So, the answer is option (C).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The wavelength of a photon is $5000\mathring {A} $.Its momentum will be:

  1. $1.32 \times 10^{-27 }kg\times$ meter/sec
  2. $1.5 \times 10^{-27 }kg\times$ meter/sec
  3. $2.32 \times 10^{-27 }kg\times$ meter/sec
  4. $5 \times 10^{-27 }kg\times$ meter/sec
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Momentum p = h/lambda = (6.6 * 10^-34) / (5000 * 10^-10) = 1.32 * 10^-27 kg m/s.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Momentum of $ \gamma -ray $ photon of energy 3 KeV in kg-m/s will be 

  1. $ 1.6 \times 10^{-19} $
  2. $ 1.6 \times 10^{-2} $
  3. $ 1.6 \times 10^{-24} $
  4. $ 1.6 \times 10^{-27} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Energy E = 3 KeV = 3000 * 1.6 * 10^-19 J = 4.8 * 10^-16 J. Momentum p = E/c = (4.8 * 10^-16) / (3 * 10^8) = 1.6 * 10^-24 kg m/s.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

What is the momentum of a photon having frequency $1.5 \times 10^{13} Hz $ 

  1. $3.3 \times 10 - 29\, kg\,m/s $
  2. $3.3 \times 10 - 34\, kg\, m/s$
  3. $6.6 \times 10 - 34\, kg\, m/s$
  4. $6.6 \times 10 - 30\, kg\, m/s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Momentum p = E/c = hf/c = (6.6 * 10^-34 * 1.5 * 10^13) / (3 * 10^8) = 3.3 * 10^-29 kg m/s.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The ratio of de-Broglie wavelengths of proton and $\alpha$-particle having same kinetic energy is

  1. $\sqrt2 :1$
  2. $2\sqrt2 :1$

  3. 2 :1

  4. 4 : 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

De Broglie wavelength is given by:


$\lambda = \dfrac{h}{p}$ 

Writing momentum as a a function of kinetic energy and mass

$\lambda = \dfrac{h}{\sqrt{2Em}}$

i.e.   $\dfrac{\lambda _{proton}}{\lambda _{alpha}} = \sqrt{\dfrac{m _{alpha}}{m _{proton}}}$

$m _{alpha} = 4m _{proton}$
So,

$\dfrac{\lambda _{proton}}{\lambda _{alpha}} = \sqrt{\dfrac{m _{alpha}}{m _{proton}}} = \sqrt{\dfrac{4}{1}} = 2$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.40$A^{0}$. Then the maximum energy of the photon of the emitted radiation. 

($h = 6.63 \times10^{-34}Js$ and $c= 3 \times10^{8}$ m/s)

  1. $4.9725\times 10^{-15}J$
  2. $6.28\times 10^{-15}J$
  3. $3.15\times 10^{-15}J$
  4. $2.98\times 10^{-15}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\lambda _{cutoff} =\ \dfrac{hc}{E}$

$E _{max}=\ \dfrac{20\times 10^{-26}}{0.4\times 10^{-10}}$

$=\ 4.9725\times 10^{-15}\ J$

So, the answer is option (A).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

When the momentum of a photon is changed by an amount $p'$ then the corresponding change in the de-Broglie wavelength is found to be $0.20$%. Then, the original momentum of the photon was

  1. $300 p'$
  2. $500 p'$
  3. $400 p'$
  4. $100 p'$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
As, we know de-Broglie wavelength,
$\lambda = \dfrac{h}{p}$
$\therefore \lambda \propto \dfrac{1}{p}$
$\Rightarrow \dfrac{\Delta p}{p} = - \dfrac{\Delta \lambda}{\lambda} \therefore \left| \dfrac{\Delta p}{p} \right | = \left| \dfrac{\Delta \lambda}{\lambda} \right |$
$\Rightarrow \dfrac{p'}{p} = \dfrac{0.20}{100} = \dfrac{1}{500}$
or, $p = 500 p'$
Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

The formula for kinetic mass of photon is:
where $h$ is Planck's constant and $v,\lambda, c$ are frequency, wavelength and speed of photon respectively.

  1. $\cfrac { hv }{ \lambda } $
  2. $\cfrac { h }{ c\lambda } $
  3. $\cfrac { hv }{ c } $
  4. $\cfrac { h\lambda }{ c } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy of the photon is given by,

$E=hv=\dfrac{hc}{\lambda}$,  ...(1)
also for particle of zero mass,
$E=mc^2$,   ...(2)
form eq(1) and (2) we get,
$mc^2=\dfrac{hc}{\lambda}$
$\therefore$ kinetic mass of photon$= m=\dfrac{h}{c\lambda}$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

If momentum of a photon of an electromagnetic radiation is $3.3\times 10^{-29}$ kg m/sec, then frequency of associated wave is:

  1. $3.0\times 10^3Hz$
  2. $6.0\times 10^3Hz$
  3. $7.5\times 10^5Hz$
  4. $1.5\times 10^{13}Hz$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p=\dfrac {hv}{c}$
$\Rightarrow v=\dfrac {cp}{h}=\dfrac {(3\times 20^8)\times (3\cdot 3\times 10{-29})}{6\cdot 6\times 10^{-34}}$


$=1\cdot 5\times 10^{13}Hz.$